`f(x) = 2csc((3x)/2), (0, 2pi)` Find the points of inflection.

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Chapter 3, 3.4 - Problem 26 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=2csc((3x)/2)`

differentiating,

`f'(x)=-2csc(3x/2)cot(3x/2)*(3/2)`

` ` `f'(x)=-3csc(3x/2)cot(3x/2)`

differentiating again,

`f''(x)=-3(csc(3x/2)d/dxcot(3x/2)+cot(3x/2)d/dxcsc(3x/2))`

`f''(x)=-3(csc(3x/2)(-csc^2(3x/2)*(3/2))+cot(3x/2)(-csc(3x/2)cot(3x/2)*(3/2))`

`f''(x)=-3(-3/2csc^3(3x/2)-3/2csc(3x/2)cot^2(3x/2))`

`f''(x)=9/2csc(3x/2)(csc^2(3x/2)+cot^2(3x/2))`

`f''(x)=9/2csc(3x/2)(1+2cot^2(3x/2))`

Now let us find out x for which f''(x)=0,

csc(3x/2)=0 , No value of x exists for which csc(3x/2)=0

`1+2cot^2(3x/2)=0`

`cot^2(3x/2)=-1/2`

So no solution exists.

So,f''(x) is undefined for x=0

Now let us test the concavity of the function by taking test value of -pi/2 and pi/2.

`f''(-pi/2)=9/2csc(-3pi/4)(1+2cot^2(-3pi/4))`

`f''(-pi/2)=9/2(-sqrt(2))(1+2)`

`f''(-pi/2)=-27/2(sqrt(2))`

`f''(pi/2)=9/2csc(3pi/4)(1+2cot^2(3pi/4))`

`f''(pi/2)=9/2(sqrt(2)(1+2)`

`f''(pi/2)=27/2sqrt(2)`

Though the concavity of the function is changing, but that does not mean that the inflection point exists at x=0, since the function is not defined at x=0.

Therefore the inflection points do not exist for the function.

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